Meriam, Kraige & Bolton — Rigid Bodies: Equivalent Systems of Forces: Problem 3_5

⚡ Mecademy AIENG정역학 · ch3  Problem Statement Determine the force required to maintain the 200-kg engine in the position for which . The diameter of the pulley at is negligible. Problem 3/5 (a) Determination of Force for Equilibrium ● Calculation Process 1. Formula: From the equilibrium of the joint at point , the sum of forces in the horizontal () and vertical () directions must be zero: where is the weight of the engine, is the tension in cable , and is the angle of cable with the horizontal. 2. Substitution: From the geometry, triangle is isosceles because and . Given vertex angle , the base angles are . Thus, . Engine weight: . Equilibrium equations: 3. Calculation: Substitute the expression for into the vertical equilibrium equation: Pθ= 30 ∘ B P C xy F = ∑ x 0⟹Pcosφ−T cosθ= AC 0 F = ∑ y 0⟹Psinφ+T sinθ− AC W=0 W=mgT AC ACφ BC ABCAB=2 m AC=2 mθ=30 ∘ ∠ABC= ∠ACB= = 2 180−30 ∘∘ 75 ∘ φ=75 ∘ W=200 kg× 9.81 m/s= 2 1962 N Pcos75− ∘ T cos30= AC ∘ 0⟹T = AC P cos30 ∘ cos75 ∘ Psin75+ ∘ Tsin30− AC ∘ 1962=0 T AC Psin75+ ∘ P sin30= ( cos30 ∘ cos75 ∘ ) ∘ 1962 Psin75+cos75tan30=[ ∘∘∘ ]1962 Using trigonometric identities: . Since : 4. Result: ● Final Conclusion: The force required to maintain the engine in the specified equilibrium position is (or ). ✨ Final Answer Summary (a) Mecademy AI Solution · ENGProblem 3/5 sin75+ ∘ cos75tan30= ∘∘ = cos30 ∘ sin75cos30+cos75sin30 ∘∘∘∘ = cos30 ∘ sin(75+30) ∘∘ cos30 ∘ sin105 ∘ sin105= ∘ sin75 ∘ P= sin105 ∘ 1962cos30 ∘ P= ≈ 0.965926 1962×0.866025 1759.